Calculus derivitives help me please!?

(1/4) * 4 * x^3 - 3 * x^2 - 2 * 2 * x - 0 =>

x^3 - 3x^2 - 4x =>

x * (x^2 - 3x - 4) =>

x * (x - 4) * (x + 1)

Set that to 0 and solve

x^4 + x^3 - 3x^2 - x + 1

4x^3 + 3x^2 - 6x - 1

12x^2 + 6x - 6

6 * (2x^2 + x - 1)

Set that to 0 and solve

X^4+x^3-3x^2-x+1

d/dx(x^4+x^3-3x^2-x+1)

= 4x^3+3x^2-6x-1 = 0

The point of inflection is the solution of the above equation

x=1 is a sulution

x= -1.5931 is a solution

x =-1.5693 is s solution

((www.wolframalpha.com)

you may desire to prepare the chain rule. This function has 5 nested applications: cos, sqrt, sin, tan, and a linear function (?x). bear in mind that cos' = -sin, sqrt' = a million / (2*sqrt), sin' = cos, tan' = sec^2 or a million + tan^2, and (ax)' = a y = cos?(sin(tan?x)) y' = -sin?(sin(tan?x)) * (?(sin(tan?x)))' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * (sin(tan?x))' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * cos(tan?x) * (tan?x)' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * cos(tan?x) * (a million + tan^2(?x)) * (?x)' = -sin?(sin(tan?x)) * (a million / 2?(sin(tan?x))) * cos(tan?x) * (a million + tan^2(?x)) * ? I flow away any needed simplication to you, yet I do see a ?(sin(tan?x)) in the two the numerator and denominator.